Question

in a certain algebra 2 class of 28 students, 21 of them play basketball and 20 of them play baseball. there are 5 students who play neither sport. what is the probability that a student chosen randomly from the class plays both basketball and baseball?

Explanation:

Step1: Find the number of students who play at least one sport

Total students \(n(T)=28\), students who play neither \(n(N) = 5\).
Number of students who play at least one sport \(n(A\cup B)=n(T)-n(N)=28 - 5=23\).

Step2: Use the formula \(n(A\cup B)=n(A)+n(B)-n(A\cap B)\)

Let \(n(A) = 21\) (basketball players), \(n(B)=20\) (baseball players).
Substitute into the formula: \(23=21 + 20-n(A\cap B)\).
Solve for \(n(A\cap B)\):

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Step3: Calculate the probability

Probability \(P=\frac{n(A\cap B)}{n(T)}\).
Substitute \(n(A\cap B) = 18\) and \(n(T)=28\): \(P=\frac{18}{28}=\frac{9}{14}\).

Answer:

\(\frac{9}{14}\)